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\title{\heiti\zihao{2} 习题15.1}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{证明:若函数 $f(\mathbf{x})$ 在点 $\mathbf{x}_0$ 可微,则函数在点 $\mathbf{x}_0$ 连续且各个偏导数存在.}
\begin{proof}
	记 $\Delta \mathbf{x}=\mathbf{x}-\mathbf{x}_{0}$, 则 $\mathbf{x} \rightarrow \mathbf{x}_{0}$ 等价于 $\Delta \mathbf{x} \rightarrow 0$, 即其等价于: 对于 $\forall i(1 \leqslant i \leqslant n)$, 有 $\Delta \mathbf{x}_{i} \rightarrow 0 .$ 由此我们有
	$$
		\lim _{\Delta \mathbf{x} \rightarrow 0}\left[f\left(\mathbf{x}_{0}+\Delta \mathbf{x}\right)-f\left(\mathbf{x}_{0}\right)\right]=\lim _{\Delta \mathbf{x} \rightarrow 0}\left[\sum_{i=1}^{n} A_{i} \Delta x_i+o(|\Delta \mathbf{x}|)\right]=0
	$$
	即 $f(\mathbf{x})$ 在 $\mathbf{x}_{0}$ 处连续.\par
	对任何固定的 $i(1 \leqslant i \leqslant n)$, 当 $j=1,2, \cdots, n$ 且 $j \neq i$ 时， 令 $x_{j}=x_{j}^{0}$, 即 $\Delta x_{j}=0 .$ 此时我们有 $\Delta x=\left(0, \cdots, \Delta x_{i}, \cdots, 0\right)$ 和
	$|\Delta x|=\left|\Delta x_{i}\right|$, 因此有
	$$
		\begin{aligned}
			\lim_{\Delta x_{i} \rightarrow 0} \dfrac{f\left(x_{1}^{0}, x_{2}^{0}, \cdots, x_{i-1}^{0}, x_{i}^{0}+\Delta x_{i}, x_{i+1}^{0}, \cdots, x_{n}^{0}\right)-f\left(x_{0}\right)}{\Delta x_{i}}
			=\lim _{\Delta x_{i} \rightarrow 0}\left(A_{i}+d\dfrac{o\left(\left|\Delta x_{i}\right|\right)}{\Delta x_{i}}\right)=A_{i}
		\end{aligned}
	$$
	从而$f(\mathbf{x})$关于$x_i$可偏导,且$\dfrac{\partial f(\mathbf{x}_0)}{\mathbf{x}_i}=A_i$
\end{proof}

\section{求下列函数的偏导数}
\subsection{$z=x^{2} \cos y$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial z}{\partial x}=2 x \cos y    \\
		 & \dfrac{\partial z}{\partial y}=-x^{2} \sin y
	\end{aligned}
$$

\subsection{$z=x^{y}$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial z}{\partial x}=y x^{y-1}   \\
		 & \dfrac{\partial z}{\partial y}=x^{y} \ln x
	\end{aligned}
$$

\subsection{$u=\dfrac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial u}{\partial x}=-\dfrac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} \\
		 & \dfrac{\partial u}{\partial y}=-\dfrac{y}{\sqrt{x^{2}+y^{2}+z^{2}}} \\
		 & \dfrac{\partial u}{\partial z}=-\dfrac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}
	\end{aligned}
$$

\subsection{$z=\ln \tan \dfrac{x}{y}$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial z}{\partial x}=\dfrac{-x}{y^{2} \cos \dfrac{x}{y} \sin \dfrac{x}{y}} \\
		 & \dfrac{\partial z}{\partial y}=\dfrac{y}{y^{2} \cos \dfrac{x}{y} \sin \dfrac{x}{y}}
	\end{aligned}
$$

\subsection{$z=\dfrac{x}{\sqrt{x^{2}+y^{2}}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial z}{\partial x}=\dfrac{y^{2}}{\sqrt{x^{2}+y^{2}}} \\
		 & \dfrac{\partial z}{\partial y}=\dfrac{-x y}{\sqrt{x^{2}+y^{2}}}
	\end{aligned}
$$

\subsection{$z=y^{\ln x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & \dfrac{\partial z}{\partial x}=\dfrac{y^{\ln x} \ln y}{x} \\
		 & \dfrac{\partial z}{\partial y}=y^{\ln x-1} \ln x
	\end{aligned}
$$

\section{求下列函数的全微分}
\subsection{$z=3 x^{2} y+\dfrac{x}{y} $}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial z}{\partial x} = 6xy+\dfrac{1}{y}                                       \\
		           & \dfrac{\partial z}{\partial y} = 3x^2-\dfrac{x}{y^2}                                    \\
		\therefore & \mathrm{d} z=\left(6xy+\dfrac{1}{y}\right)\Delta x+\left(3x^2-\dfrac{x}{y^2}\right)\Delta y
	\end{aligned}
$$

\subsection{$z=\ln \left(2+x^{2}+y^{2}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial z}{\partial x} = \dfrac{2x}{2+x^2+y^2}                                           \\
		           & \dfrac{\partial z}{\partial y} = \dfrac{2y}{2+x^2+y^2}                                           \\
		\therefore & \mathrm{d} z = \left(\dfrac{2x}{2+x^2+y^2}\right)\Delta x+\left(\dfrac{2y}{2+x^2+y^2}\right)\Delta y
	\end{aligned}
$$

\subsection{$z=\dfrac{x+y}{x-y} $}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial z}{\partial x} = -\dfrac{2y}{(x-y)^2}                                         \\
		           & \dfrac{\partial z}{\partial y} = \dfrac{2x}{(x-y)^2}                                          \\
		\therefore & \mathrm{d} z = \left(-\dfrac{2y}{(x-y)^2}\right)\Delta x+\left(\dfrac{2x}{(x-y)^2}\right)\Delta y
	\end{aligned}
$$

\subsection{$z=\mathrm{e}^{x} \sin (x+y)$}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial z}{\partial x} = \mathrm{e}^{x}\sin(x+y)+\mathrm{e}^{x}\cos (x + y)                                             \\
		           & \dfrac{\partial z}{\partial y} = \mathrm{e}^{x}\cos(x+y)                                                                        \\
		\therefore & \mathrm{d} z = \left(\mathrm{e}^{x}\sin(x+y)+\mathrm{e}^{x}\cos (x + y)\right)\Delta x+\left(\mathrm{e}^{x}\cos(x+y)\right)\Delta y
	\end{aligned}
$$

\subsection{$u=\sqrt{x^{2}+y^{2}+z^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial u}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2+z^2}}                                                                                                      \\
		           & \dfrac{\partial u}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2+z^2}}                                                                                                      \\
		           & \dfrac{\partial u}{\partial z} = \dfrac{z}{\sqrt{x^2+y^2+z^2}}                                                                                                      \\
		\therefore & \mathrm{d} u = \left(\dfrac{x}{\sqrt{x^2+y^2+z^2}}\right)\Delta x+\left(\dfrac{y}{\sqrt{x^2+y^2+z^2}}\right)\Delta y+\left(\dfrac{z}{\sqrt{x^2+y^2+z^2}}\right)\Delta z
	\end{aligned}
$$

\subsection{$u=x+\sin \dfrac{y}{2}+e^{y z}$}
\textbf{解}\quad
$$
	\begin{aligned}
		           & \dfrac{\partial u}{\partial x} = 1                                                                                 \\
		           & \dfrac{\partial u}{\partial y} = \dfrac{\cos\dfrac{y}{2}}{2}+z\mathrm{e}^{yz}                                      \\
		           & \dfrac{\partial u}{\partial z} = ye^{yz}                                                                           \\
		\therefore & \mathrm{d} u = \Delta x+\left(\dfrac{\cos\dfrac{y}{2}}{2}+z\mathrm{e}^{yz}\right)\Delta y+\left(ye^{yz}\right)\Delta z
	\end{aligned}
$$

\section{计算全增量与全微分}
\subsection{求函数 $z=x^{2} y^{2}$ 在点 $(2,-1)$ 处,当 $\Delta x=0.02, \Delta y=-0.01$ 时的全增量与全微分}
\textbf{解}\quad
$$
	\begin{aligned}
		\Delta z_0                     & = f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0) \sim 0.16               \\
		\dfrac{\partial z}{\partial x} & =2xy^2                                                            \\
		\dfrac{\partial z}{\partial y} & =2yx^2                                                            \\
		\therefore \mathrm{d} z            & \sim \left(2xy^2\right)\Delta x+\left(2yx^2\right)\Delta y = 0.12
	\end{aligned}
$$

\subsection{求 $z=\ln (x y)$ 在点 $(2,1)$ 的全微分}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =\dfrac{1}{x}                             \\
		\dfrac{\partial z}{\partial y} & =\dfrac{1}{y}                             \\
		\therefore \mathrm{d} z            & = \dfrac{\Delta x}{x}+\dfrac{\Delta y}{y} \\
		                               & = \dfrac{\Delta x}{2}+\Delta y
	\end{aligned}
$$

\subsection{$u=\cos \left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)$ 在点 $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ 处的全微分}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x_i} & =-2x_i\sin\left(\sum_{i=1}^{n}x_i\right)                                       \\
		\therefore \mathrm{d} u              & = \sum_{i=1}^{n}\left(-2x_i\sin\left(\sum_{i=1}^{n}x_i\right)\right)\Delta x_i
	\end{aligned}
$$

\section{判断函数$$f(x, y)=\left\{\begin{array}{cl}\dfrac{2 x y^{3}}{x^{2}+y^{4}}, & x^{2}+y^{2} \neq 0 \\0, & x^{2}+y^{2}=0\end{array}\right.$$在 $(0,0)$ 点的可微性.}
\textbf{解}$1^\circ$\quad
由定义:
$$
	\begin{aligned}
		\mathrm{d} f = \dfrac{2\Delta x \left(\Delta y\right)^3}{\left(\left(\Delta x\right)^2+\left(\Delta y\right)^4\right)\sqrt{\left(\Delta x\right)^2+\left(\Delta y\right)^2}}
	\end{aligned}
$$
分别令$\Delta x = \Delta y$和$\Delta x = (\Delta y)^2$,可分别解得极限为$0,1$,可见$\lim\limits_{(x,y)\rightarrow (0,0)}\dfrac{f(x,y)-f(0,0)}{||(x,y)||}$极限不存在,所以在$(0,0)$处不可微.\par 
\textbf{解}$2^\circ$\quad
$$
	\begin{aligned}
		\mathrm{d} f &= \lim_{r\rightarrow 0}\dfrac{2r^4\cos \theta \sin^3 \theta}{r^3\left(\cos^2 \theta+r^2\sin^4\theta\right)}\\
		&=\lim_{r\rightarrow 0}\dfrac{2r \sin^3 \theta}{\cos \theta}
	\end{aligned}
$$
其关于$\theta$不一致,所以不可微.\par 
关于$\theta$不一致:显然取$\theta = \dfrac{\pi}{2}-r$,从而$\lim\limits_{r\rightarrow 0}\dfrac{2r \sin^3 \theta}{\cos \theta}=2$,再取$\theta = \dfrac{\pi}{4}$,得到极限为$0$,可见极限与收敛路径有关.
\section{求 $f(x, y, z)=x^{2}+y^{2}+z^{2}$ 在点 $(1,1,1)$ 沿方向 $l:(2,1,2)$ 的方向导数.}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial f}{\partial x}            & =2x                                                                       \\
		\dfrac{\partial f}{\partial y}            & =2y                                                                       \\
		\dfrac{\partial f}{\partial z}            & =2z                                                                       \\
		\cos\theta_x                              & = \dfrac{1}{\sqrt{2}},\cos\theta_y = 0,\cos\theta_z = \dfrac{1}{\sqrt{2}} \\
		\therefore \dfrac{\partial f}{\partial v} & =\dfrac{2}{\sqrt{2}}x+\dfrac{2}{\sqrt{2}}z = 2\sqrt{2}
	\end{aligned}
$$

\section{计算 $\sqrt{(1.02)^{3}+(1.97)^{3}}$ 的近似值.}
\textbf{解}\quad
记$z=\sqrt{x^2+y^2}$,且有
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =\dfrac{3x^2}{2\sqrt{x^3+y^3}}                                                \\
		\dfrac{\partial z}{\partial y} & =\dfrac{3y^2}{2\sqrt{x^3+y^3}}                                                \\
		\therefore \Delta z            & \sim \dfrac{3x^2}{2\sqrt{x^3+y^3}}\Delta x+\dfrac{3y^2}{2\sqrt{x^3+y^3}}\Delta y \\
		                               & = -0.05
	\end{aligned}
$$
所以$\sqrt{(1.02)^{3}+(1.97)^{3}} \sim 3-0.05=2.95$.


\section{设函数 $u=\ln \left(\dfrac{1}{r}\right)$, 其中 $r=\sqrt{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}$, 求 $u$ 的梯度.}
\textbf{解}\quad
$$
	\begin{aligned}
        \dfrac{\partial u}{\partial x} & =\dfrac{a-x}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}\\
        \dfrac{\partial u}{\partial y} & =\dfrac{b-y}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}\\
        \dfrac{\partial u}{\partial z} & =\dfrac{c-z}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}
	\end{aligned}
$$
从而
$$
\nabla u=\left(\dfrac{a-x}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}},\dfrac{b-y}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}},\dfrac{c-z}{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}\right)
$$

\section{证明:函数$$f(x, y)=\left\{\begin{array}{cc}\dfrac{x y}{\sqrt{x^{2}+y^{2}}}, & x^{2}+y^{2} \neq 0 \\0, & x^{2}+y^{2}=0\end{array}\right.$$在 $(0,0)$ 点连续、可偏导,但不可微.}
\begin{proof}
连续性:只需要证明$\lim\limits_{(x,y)\rightarrow(0,0)}\dfrac{x y}{\sqrt{x^{2}+y^{2}}}=0$即可.由均值不等式:
$$
\lim\limits_{(x,y)\rightarrow(0,0)}\left|\dfrac{x y}{\sqrt{x^{2}+y^{2}}}\right|\leqslant\lim\limits_{(x,y)\rightarrow(0,0)}\sqrt{\dfrac{xy}{2}}=0
$$
显然在$(0,0)$处连续.\par 
在$(0,0)$处,有
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x}&=\dfrac{0-0}{x-0}=0\\
    \dfrac{\partial f}{\partial y}&=\dfrac{0-0}{y-0}=0
\end{aligned}
$$
显然$f$可偏导.\par 
但由于
$$
\begin{aligned}
    \lim\limits_{\mathbf{x}\rightarrow 0}\dfrac{f(\mathbf{x}+\Delta \mathbf{x})-f(\mathbf{x})}{\mathbf{x}-0}&=\lim\limits_{(\Delta x,\Delta y)\rightarrow(0,0)}\dfrac{\Delta x\Delta y}{\Delta x^2+\Delta y^2}=\lim\limits_{r\rightarrow 0}\sin \theta \cos \theta
\end{aligned}
$$
其与$\theta$的取值有关,即与收敛路径有关,从而重极限不存在,所以不可微.
\end{proof}

\section{证明:函数$f(x, y)=\left\{\begin{array}{cl}\dfrac{2 x y^{2}}{x^{2}+y^{4}}, & x^{2}+y^{2} \neq 0 \\ 0, & x^{2}+y^{2}=0\end{array}\right.$在$(0,0)$点沿各个方向的方向导数都存在,但函数在$(0,0)$点既不连续也不可微.}
\begin{proof}
先证明其不连续且不可微.令$x = X$,并记$y^2 = Y$,从而有
$$
\lim_{(X,Y)\rightarrow(0,0)}\dfrac{2XY}{X^2+Y^2}=\lim_{r\rightarrow 0}2\sin \theta\cos \theta
$$
其与收敛路径有关,从而不连续.由于其不连续,从而不可微.
偏导数都存在,有
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x} &= \lim_{\Delta x\rightarrow 0}\dfrac{0}{\Delta x} = 0\\
	\dfrac{\partial f}{\partial y} &= \lim_{\Delta y\rightarrow 0}\dfrac{0}{\Delta y} = 0
\end{aligned}
$$
设$\mathbf{v}$的方向为$y=kx$,则
$$
\begin{aligned}
	\dfrac{\partial f}{\partial \mathbf{v}}&=\lim_{(\Delta x,\Delta y)\rightarrow(0,0)}\dfrac{\dfrac{2 \Delta x \Delta y^{2}}{\Delta x^{2}+\Delta y^{4}}-0}{||(\Delta x,\Delta y)||}\\
	&=\lim_{(\Delta x,\Delta y)\rightarrow(0,0)}\dfrac{2 \Delta x \Delta y^{2}}{(\Delta x^{2}+\Delta y^{4})\sqrt{\Delta x^2+\Delta y^2}}\\
	&=dfrac{2k^2}{\sqrt{1+k^2}}\\
	&=\infty
\end{aligned}
$$
从而任意方向导数都存在.\par 
或者
$$
\begin{aligned}
	\dfrac{\partial f}{\partial \mathbf{v}}&=\lim_{r\rightarrow 0}\dfrac{2\cos \theta\sin^2\theta}{\cos \theta r^2\sin^4\theta} = 2\sin^2\theta
\end{aligned}
$$
所以任意方向导数都存在.
\end{proof}








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